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Genrish500 [490]
3 years ago
13

What is the Y intercept of a linear equation written in slope intercept form y=Mx+b

Mathematics
1 answer:
DochEvi [55]3 years ago
5 0
The y-intercept / y-axis
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rob is asked to graph this system of equations : 6 x +2y = 5 y = 3x =2 how many times will they intersect
ella [17]

Answer:

The lines will not intersect.

Step-by-step explanation:

I took the test and this was the correct answer, there isnt a solution.

4 0
3 years ago
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Mario has nine Green, eight red, 10 brown, six orange, and nine blue M&Ms what fraction of the M&Ms are orange
gizmo_the_mogwai [7]

6 orange. 42 total. 6/42 = 1/7 of the M&Ms are orange.

6 0
3 years ago
What is 8q+6;q=2 its expression for the given value of the variable
Leya [2.2K]

Answer:

8q + 6 = 8*2 + 6 = 16 + 6 = 22




5 0
3 years ago
Please help me with this question​
Trava [24]

Applying the formula for the area of a sector and length of an arc, the value of k is calculated as: 27.

<h3>What is the Area of a Sector?</h3>

Area of a sector of a circle = ∅/360 × πr²

<h3>What is the Length of an Arc?</h3>

Length of arc = ∅/360 × 2πr

Given the following:

  • Radius (r) = 9 cm
  • Length of arc = 6π cm
  • Area of sector = kπ cm²

Find ∅ of the sector using the formula for length of acr:

∅/360 × 2πr = 6π

Plug in the value of r

∅/360 × 2π(9) = 6π

∅/360 × 18π = 6π

Divide both sides by 18π

∅/360 = 6π/18π

∅/360 = 1/3

Multiply both sides by 360

∅ = 1/3 × 360

∅ = 120°

Find the area of the sector:

Area = ∅/360 × πr² = 120/360 × π(9²)

Area = 1/3 × π81

Area = 27π

Therefore, the value of k is 27.

Learn more about area of sector on:

brainly.com/question/22972014

4 0
2 years ago
Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
Radda [10]

Answer:

{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  \dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes arey = k \pm \dfrac{b}{a}(x - h)

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x

Thus,  asymptote has the slope of

\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0
3 years ago
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